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liq [111]
3 years ago
12

Polygon ABCDE is reflected across the x-axis to form polygon A′B′C′D′E′. Then polygon A′B′C′D′E′ is dilated by a scale factor of

0.5 about D′ to form polygon LMNOP. Vertices A and B correspond to L and M, respectively.
Mathematics
1 answer:
PIT_PIT [208]3 years ago
6 0
Hope this helps 
The vertices<span> of </span>polygon ABCD<span> are at A(1, 1), </span>B<span>(2, 3), </span>C<span>(3, 2), and </span>D<span>(2, 1). </span>ABCD<span> is </span>reflected across<span> the </span>x-axis<span> and translated 2 units up to </span>form polygon<span> - 914438.</span>

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A sector of a circle has central angle θ=5π/6 and area 15π ft2. Find the radius of this circle.
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MA_775_DIABLO [31]
1. Ans:(A) 123

Given function: f(x) = 8x^2 + 11x
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)
=> \frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11
=> \frac{d}{dx} f(x) = 16x + 11

Now at x = 7:
\frac{d}{dx} f(7) = 16(7) + 11

=> \frac{d}{dx} f(7) = 123

2. Ans:(B) 3

Given function: f(x) =3x + 8
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)
=> \frac{d}{dx} f(x) = 3*1 + 0
=> \frac{d}{dx} f(x) = 3

Now at x = 4:
\frac{d}{dx} f(4) = 3 (as constant)

=>Ans:  \frac{d}{dx} f(4) = 3

3. Ans:(D) -5

Given function: f(x) = \frac{5}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})
=> \frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = -5x^{-2}

Now at x = -1:
\frac{d}{dx} f(-1) = -5(-1)^{-2}

=> \frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}
=> Ans: \frac{d}{dx} f(-1) = -5

4. Ans:(C) 7 divided by 9

Given function: f(x) = \frac{-7}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})
=> \frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = 7x^{-2}

Now at x = -3:
\frac{d}{dx} f(-3) = 7(-3)^{-2}

=> \frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}
=> Ans: \frac{d}{dx} f(-3) = \frac{7}{9}

5. Ans:(C) -8

Given function: 
f(x) = x^2 - 8

Now if we apply limit:
\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)

=> \lim_{x \to 0} f(x) = (0)^2 - 8
=> Ans: \lim_{x \to 0} f(x) = - 8

6. Ans:(C) 9

Given function: 
f(x) = x^2 + 3x - 1

Now if we apply limit:
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)

=> \lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1
=> Ans: \lim_{x \to 2} f(x) = 4 + 6 - 1 = 9

7. Ans:(D) doesn't exist.

Given function: f(x) = -6 + \frac{x}{x^4}
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: f(x) = 9 + \frac{x}{x^3}
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: \lim_{x \to 9} f(x) where,
f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.

If both sides are equal on applying limit then limit does exist.

Let check:
If x \textless 9: answer would be 9+9 = 18
If x \geq 9: answer would be 9-9 = 0

Since both are not equal, as 18 \neq 0, hence limit doesn't exist.


12. Ans:(B) Limit doesn't exist.

Find out: \lim_{x \to 1} f(x) where,

f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 \neq 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x                      f(x)=1/(x-9)       

----------------------------------------

8.9                       -10

8.99                     -100

8.999                   -1000

8.9999                 -10000

9.0                        -∞


Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞,  x =9 (correct)

 14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = \frac{ds(t)}{dt}

Therefore,

\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6

At t=2,

Inst. velocity = -6


15. Ans: +∞,  x =7 

f(x) = 1/(x-7)^2.

Table:

x              f(x)= 1/(x-7)^2     

--------------------------

6.9             +100

6.99           +10000

6.999         +1000000

6.9999       +100000000

7.0              +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞,  x =7 (correct)

-i

7 0
3 years ago
Read 2 more answers
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